1 ) A compound gives a mass spectrum with peaks at m/z = 77 (40%), 112 (100%), 114 (33%), and essentially no other peaks. Identify the compound. First, your molecular ion peak is 112 and you have a M+2 peak at 114. Therefore, you have a halogen.

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So molecular ion peak at 136 and M+2 peak at 138, so halogen present.

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1) A compound gives a mass spectrum with peaks at m/z = 77 (40%), 112 (100%), 114 (33%), and essentially no other peaks. Identify the compound. First, your molecular ion peak is 112 and you have a M+2 peak at 114. Therefore, you have a halogen. Now, your molecular ion peak and M+2 peak are in a 3 to 1 ratio. This means chlorine. So, 112-35=77 # C’s 77/12=6 carbons so C6H5Cl. DOUS (2(6)+2-5-1)/2=4

2) While organizing the undergraduate stockroom, a new chemistry professor found a half-gallon jug containing a cloudy liquid (bp 100–105 °C), marked only "STUDENT PREP". She ran a quick mass spectrum, which is shown below. As soon as she saw the spectrum (without even checking the actual mass numbers), she said, "I know what it is." What compound is the "student prep"?

3) A laboratory student added 1-bromobutane to a flask containing dry ether and Mg turnings. An exothermic reaction resulted, and the ether boiled vigorously for several minutes. Then she added acetone to the reaction mixture, and the ether boiled even more vigorously. She added dilute acid to the mixture, and separated the layers. She evaporated the ether layer, and distilled a liquid that boiled at 143 °C. GC–MS analysis of the distillate showed one major product with a few minor impurities. The mass spectrum of the major product is shown below. Show the structure of this major product.

First you can eliminate benzoic acid and benzyl alcohol because there is no –OH peak. Second you can eliminate benzaldehyde because there is no peak at 2740 cm-1 (aldehyde peak). That leaves phenylacetone and acetophenone. Acetophenone is the answer because the carbonyl peak is at 1700 cm-1 and a simple ketone like that on phenylacetone would absorb at a higher energy.

First, 2-ethynylcyclohexanone can be eliminated because there is no peak for a carbon carbon triple bond. Second, cyclohexylmethyl alcohol is eliminated because there is no –OH peak present. Acetopheone can be eliminated next because there is no peak for aromatic C-H stretches. Also the carbonyl peak would be at a higher energy like around 1800 cm-1. 2-methyl-2-cyclohexenone can be eliminated because there is no carbon carbon double bond peak. Leaving 4-ethylcyclohexanone.

The molecular ion peak is at 162 and the M+2 peak is at 164, and they are in a 1:1 ratio, therefore there is a bromine atom. 162-79=83 83/12=6 carbons and 6x12= 72 so 83-72=11 C6H11Br So 2(6)+2-11-1=2/2=1, which means 1 double bond or 1 ring. Looking at the IR, there is no C=C peak so that means a ring.

2(5)+2-12=0 so no double bonds or rings. Also there is no –OH or C=O peaks in the IR, so it has to be an ether. Looking downfield you have a triplet and a singlet. For there to be a singlet there must be only a methyl on one side of the ether. Thus giving us the following structure. Looking at this structure, it explains the presence of the multiplets for the middle two CH2s. And finally the triplet upfield is for the terminal methyl group.

First 2(6)+2-12=2/2=1 And based on the carbonyl peak in the IR we know this is our degree of unsaturation. Also we know that there must also be an ether since there is no OH peak in the IR. And based on the proton NMR we have two types of protons. One has to be connected to the ether. And for the rest to be all the same, there must be an isobutyl group.

114/12=9 carbons 114-108=6 hydrogens C9H6 2(9)+2-6= 7 degrees of unsaturation Based on the IR we know there is a carbonyl So –CH4 add O C8H2O 2(8)+2-2=16/2=8 Lets take off a C and add 12 Hs C7H14O 2(7)+2-14=2/2=1 Which accounts for the C=O